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How do we evaluate the following series:

$$\sum_{n=1}^\infty \frac1{4n^2+2n}$$

I know that it converges by the comparison test. Wolfram Alpha gives the answer $1 - \ln(2)$, but I cannot see how to get it. The Taylor series of logarithm is nowhere near this series.

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    $\begingroup$ This series is actually one minus the alternating harmonic series, which is known to converge to $\ln(2)$. $\endgroup$ – Mark McClure May 30 '13 at 17:21
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    $\begingroup$ After all these answers, I'm still the only person who's up-voted the question. $\endgroup$ – Michael Hardy May 30 '13 at 17:53
  • $\begingroup$ Thanks for the nice edits and the up-vote too! $\endgroup$ – Vishal Gupta May 31 '13 at 4:33
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    $\begingroup$ @Vishal Just a note about mathematical English: The term "series" already incorporates summation. Therefore, one speaks of "evaluating a series" rather than "summing a series". $\endgroup$ – Lord_Farin May 31 '13 at 8:58
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Rewrite the series as follows:

$$\sum_{n=1}^\infty\frac{1}{4n^2+2n}=\sum_{n=1}^\infty\frac{1}{2n(2n+1)}=\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=\sum_{m=2}^\infty\frac{(-1)^m}{m}.$$

You may now evaluate it using the Taylor series of logarithm.

Added: The third equality is justified as follows:

Write $a_N=\sum_{n=1}^N\left(\dfrac{1}{2n}-\dfrac{1}{2n+1}\right)$ and $b_M=\sum_{m=2}^M\dfrac{(-1)^m}{m}$. The sequence of partial sums $b_M$ is convergent by the Leibniz criterion. Furthermore, $a_N = b_{2N+1}$ holds for all $N\in\mathbb N$, i.e. $(a_N)_{N=1}^\infty$ is a subsequence of $(b_M)_{M=1}^\infty$. Therefore these sequences converge to the same limit, which justifies the equality.

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  • $\begingroup$ Does it not involve rearranging the series? $\endgroup$ – Vishal Gupta May 31 '13 at 4:56
  • $\begingroup$ @Vishal: Not really. See the justification I added. $\endgroup$ – Dejan Govc May 31 '13 at 8:36
  • $\begingroup$ Thanks, that is a nice trick $\endgroup$ – Vishal Gupta Jun 8 '13 at 3:52
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$$\dfrac1{4n^2+2n} = \dfrac1{2n(2n+1)} = \dfrac1{2n} - \dfrac1{2n+1}$$ Hence, \begin{align} \lim_{m \to \infty} \sum_{n=1}^{m} \dfrac1{4n^2+2n} & = \lim_{m \to \infty} \sum_{n=1}^{m} \left(\dfrac1{2n} - \dfrac1{2n+1} \right)\\ & = \lim_{m \to \infty}\left(\dfrac12 - \dfrac13 + \dfrac14 - \dfrac15 \pm \cdots +\dfrac1{2m} - \dfrac1{2m+1}\right)\\ & = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} + \dfrac1{2m+1}\right)\\ & = 1 - \lim_{m \to \infty}\left(1-\dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 \mp \cdots -\dfrac1{2m} \right) - \lim_{m \to \infty}\dfrac1{2m+1}\\ & = 1 - \lim_{m \to \infty}\dfrac1{2m+1} - \lim_{m \to \infty} \sum_{n=1}^{2m} \dfrac{(-1)^{n-1}}{n}\\ & = 1 - 0 - \log(1+1)\,\,\, \left\{\text{Since $\log(1+x) = -\displaystyle \sum_{n=1}^{\infty}\dfrac{(-x)^{n}}n$} \right\}\\ & = 1 - \log(2) \end{align}

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Hint: $\frac 1{4n^2+2n} = \frac 1{2n}-\frac 1{2n+1}$

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Knowing the answer is helpful. Recall that $$\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots.$$ Thus $$\ln 2=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\cdots.$$ The "typical" term $\dfrac{1}{2k}-\dfrac{1}{2k+1}$ simplifies to $\dfrac{1}{2k(2k+1)}$.

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  • $\begingroup$ Are you not rearranging the series? $\endgroup$ – Vishal Gupta May 31 '13 at 4:57
  • $\begingroup$ @Vishal He's actually taking a subsequence of the partial sums. That is OK. $\endgroup$ – Pedro Tamaroff♦ Sep 3 '13 at 22:07
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$\gamma\,$: Euler-Mascheroni constant. $\quad\Psi\left(z\right)\,$: Digamma function.

\begin{align} &\\[3mm] \sum_{n = 1}^{\infty}{1 \over 4n^{2} + 2n} &= {1 \over 4}\sum_{n = 1}^{\infty}{1 \over n\left(n + 1/2\right)} = {1 \over 4}\sum_{n = 0}^{\infty}{1 \over \left(n + 1\right)\left(n + 3/2\right)} = {1 \over 4}{\Psi\left(3/2\right) - \Psi\left(1\right) \over 3/2 - 1} \\[3mm]&= {1 \over 2} \left\lbrack\Psi\left(3 \over 2\right) - \Psi\left(1\right)\right\rbrack \\[5mm]& \end{align}

$$ \Psi\left(3 \over 2\right) = \underbrace{\Psi\left(1 \over 2\right)} _{-\gamma\ -\ 2\ln\left(2\right)} + {1 \over \left(1/2\right)} = -\gamma + 2\left\lbrack 1 - \ln\left(2\right)\right\rbrack\,, \qquad\qquad \Psi\left(1\right) = -\gamma $$

$$ \begin{array}{|c|}\hline\\ {\large\quad\sum_{n = 1}^{\infty}{1 \over 4n^{2} + 2n} = 1 - \ln\left(2\right)\quad} \\ \\ \hline \end{array} $$

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  • $\begingroup$ How is $\Psi$ defined? $\endgroup$ – apnorton Sep 3 '13 at 22:00
  • $\begingroup$ @anorton $$ \Psi\left(z\right) \equiv {{\rm d}\ln\Gamma\left(z\right) \over {\rm d}z} $$ where $\Gamma\left(z\right)$ is the Gamma function. $\endgroup$ – Felix Marin Sep 3 '13 at 23:41
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Hint: Another approach is to Complete the square in the denominator and use the Fourier transform and Poison summation formula approach.

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